The Meaning of Eigenvalue Equations in Quantum Mechanics
Definition
Matrix
$n\times n$ Let us assume a matrix $A$ is given. We call $a$, which satisfies the following equation, the eigenvalue. A non-zero $n\times 1$ vector $\mathbf{x}$ corresponding to $a$ is called an eigen vector.
$$ A \mathbf{x} = a \mathbf{x} \tag{1} $$
Operator
Let us assume an operator $A$ is given. We call $a$, which satisfies the following equation, eigenvalue, () non-zero $\ket{a}$, the eigenfunction corresponding to $a$.
$$ A \ket{a} = a \ket{a} \tag{2} $$
Specifically, the equation $(2)$ is called the eigenvalue equation, and solving the eigenvalue equation for a given operator $A$ to find eigenvalues and eigenfunctions is called the eigenvalue problem.
Explanation
In physics, the term eigenfunction or eigen state is more commonly used than eigen vector. Using the ket notation, the $\ket{a}$ is written and read as [ket A]. This notation is called the Dirac notation.
Mathematical Properties
If $\ket{a}$ is the eigenfunction corresponding to $a$ of $A$, then for any constant $c$, $c\ket{a}$ is also an eigenfunction corresponding to $a$. In other words, eigenfunctions corresponding to the same eigenvalue are constant multiples of each other.
$$ \begin{align*} A(c\ket{a}) &= cA\ket{a} \\ &= ca\ket{a} \\ &= a(c\ket{a}) \end{align*} $$
Physical Interpretation
In quantum mechanics, an operator is interpreted as an act of observing a physical quantity. Thus, the eigenvalue equation below is a mathematical representation of the act of observing a physical quantity. When observing the physical quantity $A$ of the wave function $\psi$, its value can be interpreted as $a$.
$$ A \psi = a \psi $$
To make it simple, measuring a person’s weight can be likened to $A$, which is the act of stepping on the scale, $\psi$, which is the person, and $a$, which is the person’s weight. This is quite significant, as it suggests that in quantum mechanics, the physical quantities of a particle (wave) are quantized. The Hamiltonian operator $H$ is the operator observing the energy of a particle.
$$ H \psi = E \psi $$
However, if solving the eigenvalue equation for a given potential $V$ yields discrete eigenvalues like $E_{1}$, $E_{2}$, $E_{3}$, and $\dots$ instead of continuous ones, it means that the only possible energy levels for the particle under the given conditions are $E_{1}$, $E_{2}$, $E_{3}$, and $\dots$. This means that the energy is quantized. Therefore, solving the eigenvalue problem in quantum mechanics means finding the state (eigenfunction) the particle can occupy and the physical quantities (eigenvalues) it can possess in that state.
Solving the Eigenvalue Equation
Simplify $(1)$ so that the right-hand side becomes $0$
$$ \begin{align*} && A\mathbf{x} &= \lambda \mathbf{x} \\ \implies && A\mathbf{x}-\lambda \mathbf{x} &=0 \\ \implies && (A-\lambda I)\mathbf{x} &=0 \end{align*} $$
Assuming the inverse of matrix $(A-\lambda I)$ exists. Multiplying both sides by the inverse matrix,
$$ \mathbf{x}=(A-\lambda I)^{-1} \cdot 0 = 0 $$
Thus, $\mathbf{x} = 0$. However, such solutions are meaningless, so let’s focus on finding $\mathbf{x}$ instead of $0$. To do so, the inverse of $(A-\lambda I)$ must not exist. The condition for the non-existence of the inverse of any matrix is when the determinant is $0$. Thus, the condition for finding $\mathbf{x}$ instead of $0$ is as follows.
$$ |A-\lambda I |=0 $$
This equation is called the characteristic equation, and its solutions are the eigenvalues.
Finding Eigenvalues
Let us assume the matrix $A$ is given as $A=\begin{pmatrix} 6 & 2 \\ 2 & 3 \end{pmatrix}$. Then $A-\lambda I$ is as follows.
$$ \begin{pmatrix} 6 & 2 \\ 2 & 3 \end{pmatrix}-\lambda I = \begin{pmatrix} 6 -\lambda& 2 \\ 2 & 3-\lambda \end{pmatrix} $$
Solving the characteristic equation,
$$ \begin{align*} \begin{vmatrix} 6 -\lambda & 2 \\ 2 & 3-\lambda \end{vmatrix} &= (\lambda -6)(\lambda -3)-4 \\ &= \lambda^2-9\lambda+14 \\ &=(\lambda-7)(\lambda-2)=0 \end{align*} $$
Thus, we find that the two eigenvalues are $\lambda=7$ and $\lambda=2$.
Finding Eigenfunctions
Once the eigenvalues are found, the eigenfunctions corresponding to each eigenvalue can be found. Let it be $\mathbf{x}=\begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}$.
Case 1 $\lambda = 7$
$$ \begin{align*} && (A-\lambda I) \mathbf{x}&=0 \\ \implies && \begin{pmatrix} 6-7 & 2 \\ 2 & 3-7 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}&= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \implies && \begin{pmatrix} -1 & 2 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}&= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \implies && \begin{pmatrix} -1x_{1} + 2x_{2} \\ 2x_{1} -4x_{2} \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{align*} $$
In this case, $2x_{1}=x_{2}$ is satisfied by any $x_{1}$ and $x_{2}$, thus the solutions $\mathbf{x}$ can be infinitely many. However, in quantum mechanics, $\mathbf{x}$ is the wave function, and the wave function has no meaning without normalization. Therefore, any pair of $x_{1}$ and $x_{2}$ that satisfies the above equation can be chosen. Regardless of the pair chosen, the normalization process will ultimately yield the same value. The important point here is to choose the simplest form. There is no need to choose a solution that is $x_{2}=1000$ when $x_{1}=2000$. Since $x_{1}=1$ when $x_{2}=2$, the eigenfunction is
$$ \mathbf{x_{1}}=\begin{pmatrix} a \\ b \end{pmatrix}=\begin{pmatrix} 2 \\ 1 \end{pmatrix} $$
The important point here is that normalization is mandatory. The normalized eigenfunction is given as follows.
$$ \mathbf{x}_{1}=\frac{1}{\sqrt{5}}\begin{pmatrix} 2 \\ 1 \end{pmatrix} $$
Case 2 $\lambda = 2$
By the same method, the normalized eigenfunction $x_{2}$ is as follows.
$$ \mathbf{x}_{2}= \frac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ -2 \end{pmatrix} $$
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